pandas, groupby, aggregation (named)

pandas, groupby, aggregation (named).py

import pandas as pd
import numpy as np
import random

# create a dataframe
n = 50
df = pd.DataFrame({'a': random.choices(['foo', 'boo', 'bah'], k=n), 'b': np.random.rand(n), 'c': np.random.rand(n)*10})
grouped = df.groupby('a')

# example 1, one column, one function, name(s) not specified
res = grouped['b'].agg(np.mean)
print(res)

# example 2, one column, many functions, name(s) not specified
res = grouped['b'].agg([np.mean, lambda x: x.mean() - x.median()])
print(res)

# example 3, many columns, many functions, name(s) not specified
res = grouped[['b', 'c']].agg([np.mean, lambda x: x.mean() - x.median()])
print(res)

# example 4, many columns, many functions, names specified (no spaces)
res = grouped[['b', 'c']].agg(b_mean=pd.NamedAgg('b', np.mean), c_mean_median_diff=pd.NamedAgg('c', lambda x: x.mean()-x.median()))
print(res)

# example 5, many columns, many functions, names specified (spaces)
res = grouped[['b', 'c']].agg(**{'b mean': pd.NamedAgg(column='b', aggfunc=np.mean),
                                 'c mean median diff': pd.NamedAgg('c', lambda x: x.mean()-x.median())})
print(res)

# example 6, many columns, many functions, names specified (spaces)
res = grouped[['b', 'c']].agg(**{'b mean': ('b', np.mean),
                                 'c mean median diff': ('c', lambda x: x.mean()-x.median())})
print(res)